Problem: The lifespans of porcupines in a particular zoo are normally distributed. The average porcupine lives $18.8$ years; the standard deviation is $4.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a porcupine living longer than $6.2$ years.
Explanation: $18.8$ $14.6$ $23$ $10.4$ $27.2$ $6.2$ $31.4$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $18.8$ years. We know the standard deviation is $4.2$ years, so one standard deviation below the mean is $14.6$ years and one standard deviation above the mean is $23$ years. Two standard deviations below the mean is $10.4$ years and two standard deviations above the mean is $27.2$ years. Three standard deviations below the mean is $6.2$ years and three standard deviations above the mean is $31.4$ years. We are interested in the probability of a porcupine living longer than $6.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the porcupines will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the porcupines will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $6.2$ years and the other half $({0.15\%})$ will live longer than $31.4$ years. The probability of a particular porcupine living longer than $6.2$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.